How to transform a matrix in Matlab?

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Moe
Moe il 29 Feb 2016
Commentato: Priti Gujar il 15 Giu 2020
Matrix A is as follows:
A = [0 240 245 250
25 1 2 1
63 3 2 1];
I want matrix A to be transformed to B (like follows):
B = [0 240 245 250 240 245 250 240 245 250
25 1 0 1 0 1 0 0 0 0
63 0 0 1 0 1 0 1 0 0];
there are three different variables in matrix A, so, 204 to 250 (first row) in matrix B is repeated 3 times (e.g. if there were 5 variables, then 240 to 250 should be repeated 5 times). Then, value of ID = 25 has 1, so 1 is added to B(2,2). Again, A(2,3) = 2, then B(2,6) should by =1 and A(2,4)=1, then A(2,10) should be equal by =1. And same for ID#63
  4 Commenti
Moe
Moe il 29 Feb 2016
Hi Walter,
You are right, the output should be same as you wrote in your comment. I edited in main question. Do you have any idea how to get this result?
Priti Gujar
Priti Gujar il 15 Giu 2020
Use the readall function to import all the data. Check that the preprocessing function was applied to each file by plotting the Y variable as a function of Time.

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Risposte (2)

John BG
John BG il 29 Feb 2016
Hi Mohammad
MATLAB has the command linsolve to solve linear equation systems of the type A*x=b
Your question has one A:
A = [0 240 245 250; 25 1 2 1; 63 3 2 1]
and ten b:
B= [0 240 245 250 240 245 250 240 245 250;
25 1 0 0 0 1 0 0 0 1;
63 0 0 1 0 1 0 1 0 0]
One way to solve them is with a for loop:
C=zeros(4,10)
for k=1:1:10
C(:,k)=linsolve(A,B(:,k))
end
answer:
C =
Columns 1 through 6
1.00 0.00 -0.04 -0.04 -0.04 0.00
-0.00 -0.55 0.77 1.30 0.75 -0.02
0 0 0 0 0 0
0.00 1.49 0.24 -0.25 0.24 1.00
Columns 7 through 10
-0.04 -0.04 -0.04 0.00
0.78 1.27 0.77 -0.52
0 0 0 0
0.25 -0.26 0.24 1.50
test it's the correct answer
A*C
ans =
Columns 1 through 6
0.00 240.00 245.00 250.00 240.00 245.00
25.00 1.00 -0.00 -0.00 -0.00 1.00
63.00 -0.00 0 1.00 0 1.00
Columns 7 through 10
250.00 240.00 245.00 250.00
0 -0.00 -0.00 1.00
0 1.00 0 -0.00
note the type (class) has changed to double. To bring it back to, for instance, range [0 255] use uint8.
does this answer help? if so click on the thumbs-up icon link on the top of this page, thanks in advance
John
  2 Commenti
Walter Roberson
Walter Roberson il 29 Feb 2016
Mohammad Hesam comments
This is not "linsolve" problem.
John BG
John BG il 29 Feb 2016
understood, had to read the previous question to realize that A is a table and the kind of the variable tagging sought.
B has to be built by the answer, not used by it to find a transform matrix.
Thanks for mentioning Mohammad's comment.

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Andrei Bobrov
Andrei Bobrov il 29 Feb 2016
Modificato: Andrei Bobrov il 29 Feb 2016
[m,n] = size(A);
[ii,k] = ndgrid(1:n-1,1:m-1);
jj = A(2:end,2:end)';
b0 = accumarray([ii(:),jj(:),k(:)],1,[n-1,n-1,m-1]);
B = [nan,repmat(A(1,2:end),1,n-1);[A(2:end,1),reshape(b0,[],m-1)']];
or with bsxfun
[m,n] = size(A);
n1 = n - 1;
A0 = A(2:end,2:end)';
b0 = reshape( bsxfun(@eq,1:n1,reshape(A0,n1,[],m-1)),[],m-1)';
B = [nan,repmat(A(1,2:end),1,n-1);[A(2:end,1),b0]];
  2 Commenti
Moe
Moe il 29 Feb 2016
Thanks Andrei,
The output of your code is:
B = [NaN 240 245 250 240 245 250 240 245 250
25 1 0 0 0 1 1 0 0 1
63 1 0 0 0 1 0 0 0 0];
While it's different with the output that I wanted in main question. Can you edit your code?

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